package dynamicProgramming.review11_15;

public class Class115 {
//TODO:2023/11/17 给定一个字符串 s 和一个字符串 t ，计算在 s 的子序列中 t 出现的个数
    public int numDistinct(String s, String t) {
        int m = s.length();
        int n = t.length();

        int[][] dp = new int[m + 1][n + 1];
        for (int i = 0; i <= m; i++) {
            dp[i][0] = 1;
        }


        for (int i = 1; i <= m; i++) {
            for (int j = 1; j <= n; j++) {
                if (s.charAt(i - 1) == t.charAt(j - 1)) {
                    dp[i][j] = dp[i - 1][j - 1] + dp[i - 1][j];
                } else {
                    dp[i][j] = dp[i - 1][j];
                }
            }
        }

        return dp[m][n];


    }

    }
